Rank forces according to magnitude: Newton's 2nd/3rd laws

In summary: The total forces on each box are equal and opposite, so the acceleration is zero.In summary, the forces on the crates in the elevator can be ranked from largest to smallest as follows: N (Elevator on B) > N (B on A) = N (A on B) = W (Earth on A) > W (Earth on B). This ranking is based on Newton's third law, which states that the normal forces between A and B are equal and opposite, and the weight of each crate is equal to the normal force exerted on it by the Earth. Newton's second law is not relevant in this scenario, as there is no change in momentum.
  • #1
bornofflame
56
3

Homework Statement


Two crates, A and B, are in an elevator as shown. The mass of crate A is greater than the mass of crate B.

a. The elevator moves downward at constant speed.
...
iii. Rank the forces on the crate according to magnitude, from largest to smallest. Explain your reasoning, including how you used Newton's second and third laws.
p101-mech-hw-40.png
p101-mech-hw-40-fbd.png

Homework Equations



F = ma
W = mg

The Attempt at a Solution



This is what I have:
N (elevator on B) > N (B on A) = N (A on B) = W (Earth on A) = W (Earth on B)
The following ranking is valid because, per Newton's 3rd law, the normal forces between A and B are action-reaction pairs and so are equal in magnitude, which in turn means that, because A is at rest and so has a net force of zero, it's weight, W, is equal in magnitude to the normal of B on A, N (B on A).
Because B is also at rest, relative to the elevator, it's net force is also zero and so the normal force of the elevator on B, N (Elev. on B) is equal in magnitude to the forces in the opposite direction, namely N (A on B) and W (Earth on B).

I believe this is correct but my understanding of the forces is not so solid that I feel I'm 100% confident in this and I am not sure how Newton's 2nd Law plays into the explanation because except for the motion of the elevator, the boxes are not moving and even then the elevator is moving at a constant speed which, to me, speaks more to Newton's 1st law and not the 2nd.
Am I correct in my reasoning or am I missing something?
Thanks for your help.
 

Attachments

  • p101-mech-hw-40.png
    p101-mech-hw-40.png
    329 bytes · Views: 2,259
  • p101-mech-hw-40-fbd.png
    p101-mech-hw-40-fbd.png
    5.2 KB · Views: 1,431
Physics news on Phys.org
  • #2
bornofflame said:
W (Earth on A) = W (Earth on B)
How did you decide that?
 
  • Like
Likes bornofflame
  • #3
I was looking at it this way:
if N (A on B) = W (Earth on B),
N (A on B) = N (B on A),
N (B on A) = W (Earth on A)
then W (Earth on A) = W (Earth on B).

Although, now that you've pointed it out, it seems silly because of course they would both have different weights since the mass of B is less than A.

Hmmm.

So, if N (Elevator on B) = N (A on B) + W (Earth on B) then that means that N (A on B) > W (Earth on B)?
 
  • #4
bornofflame said:
if N (A on B) = W (Earth on B)
But it isn't.
bornofflame said:
So, if N (Elevator on B) = N (A on B) + W (Earth on B) then that means that N (A on B) > W (Earth on B)?
Yes.
 
  • Like
Likes bornofflame
  • #5
Ok. Thank you, very much!
So, I am pretty sure then that the final ranking is as follows:
N (Elevator on B) > N (B on A) = N (A on B) = W (Earth on A) > W (Earth on B)

Does Newton's 2nd law come into play at all in the explanation? From my understanding it's only the 1st and 3rd laws that are relevant to the explanation.
 
  • #6
bornofflame said:
Ok. Thank you, very much!
So, I am pretty sure then that the final ranking is as follows:
N (Elevator on B) > N (B on A) = N (A on B) = W (Earth on A) > W (Earth on B)

Does Newton's 2nd law come into play at all in the explanation? From my understanding it's only the 1st and 3rd laws that are relevant to the explanation.
Yes, since no momentum is changing in this question.
 
  • Like
Likes bornofflame

Related to Rank forces according to magnitude: Newton's 2nd/3rd laws

1. What is the difference between Newton's 2nd and 3rd laws?

Newton's 2nd law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. On the other hand, Newton's 3rd law states that for every action, there is an equal and opposite reaction.

2. How do you rank forces according to magnitude?

To rank forces according to magnitude, you must first calculate the net force acting on an object by adding up all the individual forces acting on it. Then, you can compare the magnitude of this net force to other forces to determine their relative rankings.

3. Can forces be negative in magnitude?

Yes, forces can have negative magnitude. This simply means that the force is acting in the opposite direction of a chosen positive direction. For example, if a force of -10 N is acting in the negative x-direction, it is equivalent to a force of 10 N acting in the positive x-direction.

4. What are some examples of forces that follow Newton's 3rd law?

Some examples of forces that follow Newton's 3rd law include a person pushing against a wall (the wall pushes back with an equal and opposite force) and a rocket launching (the force of the exhaust pushing downward causes an equal and opposite force pushing the rocket upward).

5. How does understanding Newton's 2nd and 3rd laws help in practical applications?

Understanding Newton's 2nd and 3rd laws is crucial in many practical applications, such as engineering and physics. It allows us to predict the motion of objects, design structures and machines, and calculate the forces needed to achieve certain movements or actions.

Similar threads

  • Introductory Physics Homework Help
Replies
13
Views
292
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
406
  • Introductory Physics Homework Help
Replies
3
Views
991
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
25
Views
297
  • Introductory Physics Homework Help
2
Replies
42
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
418
Back
Top